Question: A sphere is inscribed in a cube, and the cube has a surface area of 24 square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2  meters. The sphere inscribed within the cube has diameter 2 meters,  which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives \[
l^2 + l^2 + l^2 = 2^2 = 4.
\]Hence each face has surface area \[
l^2 = \frac{4}{3} \ \text{square meters}.
\]So the surface area of the inscribed cube is $6\cdot (4/3) = \boxed{8}$ square meters.